javascript 覆盖原型属性或函数
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overriding prototype property or function
提问by Mohammad Faizan khan
function Ninja(){
this.swingSword = function(){
return true;
};
}
// Should return false, but will be overridden
Ninja.prototype.swingSword = function(){
return false;
};
var ninja = new Ninja();
log( ninja.swingSword(), "Calling the instance method, not the prototype method." );
now log showing me true. which means swingSword that were defined in Ninja.prototype has overridden so how can i override the constructor function or property.?? i know that preference is given to constructor variable then why need to define a function or property inside prototype??
现在记录显示我是真的。这意味着在 Ninja.prototype 中定义的 swingSword 已被覆盖，所以我如何覆盖构造函数或属性。？我知道优先考虑构造函数变量那为什么需要在原型中定义函数或属性？
采纳答案by Stephen Kaiser
The reason to define a function on the prototype is so that it is shared between all instances. This will save you some memory rather than each instance having its own copy of a function defined in the constructor.
在原型上定义函数的原因是它在所有实例之间共享。这将为您节省一些内存，而不是每个实例都有自己的构造函数中定义的函数副本。
Some other references you might be interested in:
您可能感兴趣的其他一些参考资料：
Javascript when to use prototypes
回答by Andrew Templeton
This is by design. Do not set the value in the constructor if you want it to return false.
这是设计使然。如果您希望它返回 false，请不要在构造函数中设置该值。
You can also make a setter method:
您还可以创建一个 setter 方法：
function Ninja() {
var swordState = true;
this.swingSword = function () {
return swordState;
};
this.setSword = function (b) {
swordState = b;
};
}
// Should return false, but will be overridden
Ninja.prototype.swingSword = function () {
return false;
};
var ninja = new Ninja();
console.log(ninja.swingSword(), "Calling the instance method, not the prototype method.");
ninja.setSword(false);
console.log(ninja.swingSword()); // returns false
